Important Questions
Practice these important questions to strengthen your understanding
Question 1
Explain De Morgan’s theorems with truth tables and logic diagrams.
Answer:
De Morgan's First Theorem: \( \overline{A + B} = \overline{A} \cdot \overline{B} \)
Truth table:
| A | B | A + B | \( \overline{A + B} \) | \( \overline{A} \) | \( \overline{B} \) | \( \overline{A} \cdot \overline{B} \) |
|---|---|-------|-----------------|-------------|-------------|-------------------------|
| 0 | 0 | 0 | 1 | 1 | 1 | 1 |
| 0 | 1 | 1 | 0 | 1 | 0 | 0 |
| 1 | 0 | 1 | 0 | 0 | 1 | 0 |
| 1 | 1 | 1 | 0 | 0 | 0 | 0 |
Logic diagram: NOR gate equals bubbled AND gate (AND gate with inverted inputs).
De Morgan's Second Theorem: \( \overline{A \cdot B} = \overline{A} + \overline{B} \)
Truth table:
| A | B | \( A \cdot B \) | \( \overline{A \cdot B} \) | \( \overline{A} \) | \( \overline{B} \) | \( \overline{A} + \overline{B} \) |
|---|---|-----------|-----------------------|-------------|-------------|---------------------------|
| 0 | 0 | 0 | 1 | 1 | 1 | 1 |
| 0 | 1 | 0 | 1 | 1 | 0 | 1 |
| 1 | 0 | 0 | 1 | 0 | 1 | 1 |
| 1 | 1 | 1 | 0 | 0 | 0 | 0 |
Logic diagram: NAND gate equals bubbled OR gate (OR gate with inverted inputs).
Truth table:
| A | B | A + B | \( \overline{A + B} \) | \( \overline{A} \) | \( \overline{B} \) | \( \overline{A} \cdot \overline{B} \) |
|---|---|-------|-----------------|-------------|-------------|-------------------------|
| 0 | 0 | 0 | 1 | 1 | 1 | 1 |
| 0 | 1 | 1 | 0 | 1 | 0 | 0 |
| 1 | 0 | 1 | 0 | 0 | 1 | 0 |
| 1 | 1 | 1 | 0 | 0 | 0 | 0 |
Logic diagram: NOR gate equals bubbled AND gate (AND gate with inverted inputs).
De Morgan's Second Theorem: \( \overline{A \cdot B} = \overline{A} + \overline{B} \)
Truth table:
| A | B | \( A \cdot B \) | \( \overline{A \cdot B} \) | \( \overline{A} \) | \( \overline{B} \) | \( \overline{A} + \overline{B} \) |
|---|---|-----------|-----------------------|-------------|-------------|---------------------------|
| 0 | 0 | 0 | 1 | 1 | 1 | 1 |
| 0 | 1 | 0 | 1 | 1 | 0 | 1 |
| 1 | 0 | 0 | 1 | 0 | 1 | 1 |
| 1 | 1 | 1 | 0 | 0 | 0 | 0 |
Logic diagram: NAND gate equals bubbled OR gate (OR gate with inverted inputs).
Explanation:
De Morgan's theorems provide a way to transform expressions involving complements of sums and products into equivalent expressions involving products and sums of complements.
The first theorem states that the complement of an OR operation equals the AND of the complements.
The second theorem states that the complement of an AND operation equals the OR of the complements.
Truth tables verify that the outputs of both sides of each theorem match for all input combinations.
Logic diagrams show that a NOR gate is equivalent to an AND gate with inverted inputs and a NAND gate is equivalent to an OR gate with inverted inputs.
Understanding these theorems helps in simplifying logic circuits and designing using universal gates.
De Morgan's theorems provide a way to transform expressions involving complements of sums and products into equivalent expressions involving products and sums of complements.
The first theorem states that the complement of an OR operation equals the AND of the complements.
The second theorem states that the complement of an AND operation equals the OR of the complements.
Truth tables verify that the outputs of both sides of each theorem match for all input combinations.
Logic diagrams show that a NOR gate is equivalent to an AND gate with inverted inputs and a NAND gate is equivalent to an OR gate with inverted inputs.
Understanding these theorems helps in simplifying logic circuits and designing using universal gates.
Question 2
The manipulator in a robot is responsible for which of the following functions?
Answer:
The manipulator is responsible for performing physical tasks such as moving, grabbing, turning, and lifting objects. It acts as the robot's arm or tool to interact with the environment.
Explanation:
Manipulators include mechanical parts such as motors, pistons, grippers, wheels, and gears that enable the robot to physically interact with objects. They are controlled by the robot's controller to execute tasks requiring movement or manipulation.
Manipulators include mechanical parts such as motors, pistons, grippers, wheels, and gears that enable the robot to physically interact with objects. They are controlled by the robot's controller to execute tasks requiring movement or manipulation.
Question 3
Explain the physical meaning of Kirchhoff’s voltage rule and how it relates to conservation of energy.
Answer:
Kirchhoff’s voltage rule states that the algebraic sum of all the potential differences (voltage) around any closed loop in a circuit is zero.
This means that the total energy gained per unit charge by the charges from sources like batteries is exactly equal to the total energy lost per unit charge in the resistors or other elements in the loop.
Physically, this rule is a statement of the conservation of energy principle applied to electrical circuits.
It ensures that energy supplied by the emf sources is completely used up in overcoming resistances in the circuit, so no energy is lost or created within the loop.
This means that the total energy gained per unit charge by the charges from sources like batteries is exactly equal to the total energy lost per unit charge in the resistors or other elements in the loop.
Physically, this rule is a statement of the conservation of energy principle applied to electrical circuits.
It ensures that energy supplied by the emf sources is completely used up in overcoming resistances in the circuit, so no energy is lost or created within the loop.
Explanation:
Kirchhoff’s voltage rule is based on the fundamental principle of conservation of energy.
When a charge moves around a closed loop, it gains energy from sources like batteries and loses energy while passing through resistors.
The sum of these energy changes must be zero because energy cannot be created or destroyed.
This rule helps us analyze complex circuits by writing equations for voltage rises and drops around loops.
Teachers often explain this by comparing it to walking around a hill where the total gain and loss in height after a full circle is zero.
Kirchhoff’s voltage rule is based on the fundamental principle of conservation of energy.
When a charge moves around a closed loop, it gains energy from sources like batteries and loses energy while passing through resistors.
The sum of these energy changes must be zero because energy cannot be created or destroyed.
This rule helps us analyze complex circuits by writing equations for voltage rises and drops around loops.
Teachers often explain this by comparing it to walking around a hill where the total gain and loss in height after a full circle is zero.
Question 4
For a battery of emf \( \varepsilon \) and internal resistance \( r \) connected to an external resistance \( R \), derive the expression for the current flowing in the circuit when \( n \) identical cells are connected in series. Also, explain the condition under which the series connection of cells is advantageous.
Answer:
Total emf of \( n \) cells in series = \( n\varepsilon \)\Total internal resistance = \( nr \)\Using Ohm's law, current \( I = \) Total emf / Total resistance = \( \frac{n\varepsilon}{nr + R} \)\Series connection is advantageous when internal resistance is much smaller than external resistance (\( r \ll R \)), so \( I \approx \frac{n\varepsilon}{R} \), increasing current proportionally with \( n \).\If \( r \gg R \), current is limited by internal resistance and no advantage is gained.
Explanation:
When cells are connected in series, their emf adds up while internal resistances also add.\Ohm's law gives current as total emf divided by total resistance.\If internal resistance is negligible compared to external resistance, current increases linearly with number of cells.\This condition makes series connection beneficial.\If internal resistance dominates, current does not increase significantly, making series connection ineffective.
When cells are connected in series, their emf adds up while internal resistances also add.\Ohm's law gives current as total emf divided by total resistance.\If internal resistance is negligible compared to external resistance, current increases linearly with number of cells.\This condition makes series connection beneficial.\If internal resistance dominates, current does not increase significantly, making series connection ineffective.
Question 5
Calculate the current flowing through the galvanometer in a Wheatstone bridge if the resistors are \(P=40\,\Omega\), \(Q=80\,\Omega\), \(R=60\,\Omega\), \(S=120\,\Omega\), galvanometer resistance is 5 \(\Omega\), and battery emf is 6 V.
Answer:
Step 1 Calculate the ratio \(\frac{P}{Q} = \frac{40}{80} = \frac{1}{2}\) and \(\frac{R}{S} = \frac{60}{120} = \frac{1}{2}\). Since \(\frac{P}{Q} = \frac{R}{S}\), the Wheatstone bridge is balanced.
Step 2 In a balanced Wheatstone bridge, no current flows through the galvanometer.
Step 3 Therefore, the current through the galvanometer \(I_g = 0\).
Step 2 In a balanced Wheatstone bridge, no current flows through the galvanometer.
Step 3 Therefore, the current through the galvanometer \(I_g = 0\).
Explanation:
The Wheatstone bridge is balanced when the ratio of resistances in one branch equals the ratio in the other branch.
Here, \(\frac{P}{Q} = \frac{R}{S}\) means the bridge is balanced.
When the bridge is balanced, the potential difference across the galvanometer is zero, so no current flows through it.
This is a key property used in Wheatstone bridge measurements to determine unknown resistances accurately.
The Wheatstone bridge is balanced when the ratio of resistances in one branch equals the ratio in the other branch.
Here, \(\frac{P}{Q} = \frac{R}{S}\) means the bridge is balanced.
When the bridge is balanced, the potential difference across the galvanometer is zero, so no current flows through it.
This is a key property used in Wheatstone bridge measurements to determine unknown resistances accurately.
Question 6
Explain the role of the Internet in modern communication and mention three of its key applications.
Answer:
The Internet plays a crucial role in modern communication by connecting millions of computers worldwide, enabling fast and efficient exchange of information.
It provides a platform for various communication tools such as email, instant messaging, video calls, and social media.
Three key applications of the Internet are:
1. Personal communication through emails and social networking sites.
2. Access to information and educational resources like online courses and digital libraries.
3. Business and commerce including online shopping, banking, and remote work facilitation.
It provides a platform for various communication tools such as email, instant messaging, video calls, and social media.
Three key applications of the Internet are:
1. Personal communication through emails and social networking sites.
2. Access to information and educational resources like online courses and digital libraries.
3. Business and commerce including online shopping, banking, and remote work facilitation.
Explanation:
The Internet is fundamental in connecting people globally, making communication faster and more accessible.
It supports multiple forms of communication beyond traditional methods, enhancing personal and professional interactions.
Understanding its applications helps students appreciate the impact of technology on daily life and various industries.
Teachers can explain by showing examples of Internet use in everyday communication and business.
The Internet is fundamental in connecting people globally, making communication faster and more accessible.
It supports multiple forms of communication beyond traditional methods, enhancing personal and professional interactions.
Understanding its applications helps students appreciate the impact of technology on daily life and various industries.
Teachers can explain by showing examples of Internet use in everyday communication and business.
Question 7
Explain how the tolerance of a carbon resistor is indicated using colour bands.
Answer:
The tolerance of a carbon resistor is indicated by the fourth colour band, which is usually metallic like gold or silver. Gold represents a tolerance of ±5%, silver represents ±10%, and if there is no fourth band, the tolerance is ±20%.
Explanation:
Tolerance shows how much the actual resistance can vary from the stated value. The fourth band uses specific colours to indicate this tolerance. Gold means the resistance can vary by 5%, silver means 10%, and absence of this band means a tolerance of 20%. This helps in understanding the precision of the resistor.
Tolerance shows how much the actual resistance can vary from the stated value. The fourth band uses specific colours to indicate this tolerance. Gold means the resistance can vary by 5%, silver means 10%, and absence of this band means a tolerance of 20%. This helps in understanding the precision of the resistor.
Question 8
Discuss the challenges in interpreting Fraunhofer lines due to solar atmospheric turbulence
Answer:
Solar atmospheric turbulence causes broadening and irregularities in Fraunhofer lines.
This makes it difficult to accurately measure line positions and intensities.
Turbulence leads to Doppler broadening and shifts, complicating velocity and composition analysis.
Additionally, overlapping lines and varying atmospheric conditions add to the complexity.
Advanced modeling and high-resolution spectroscopy are required to overcome these challenges.
This makes it difficult to accurately measure line positions and intensities.
Turbulence leads to Doppler broadening and shifts, complicating velocity and composition analysis.
Additionally, overlapping lines and varying atmospheric conditions add to the complexity.
Advanced modeling and high-resolution spectroscopy are required to overcome these challenges.
Explanation:
Turbulence in the solar atmosphere causes random motions of gases, which broaden the absorption lines (Doppler broadening).
This broadening reduces the precision in determining exact wavelengths and line depths.
Irregularities from turbulence can distort line profiles, making it harder to separate effects of temperature, pressure, and magnetic fields.
Overlapping lines from different elements further complicate interpretation.
To address these issues, scientists use sophisticated instruments and computational models to simulate and correct for turbulence effects.
Turbulence in the solar atmosphere causes random motions of gases, which broaden the absorption lines (Doppler broadening).
This broadening reduces the precision in determining exact wavelengths and line depths.
Irregularities from turbulence can distort line profiles, making it harder to separate effects of temperature, pressure, and magnetic fields.
Overlapping lines from different elements further complicate interpretation.
To address these issues, scientists use sophisticated instruments and computational models to simulate and correct for turbulence effects.
Question 9
Explain the significance of the law of radioactive decay in practical applications.
Answer:
The law of radioactive decay states that the number of undecayed nuclei decreases exponentially with time.
This law helps in determining the age of archaeological and geological samples through radiocarbon dating.
It is used in medical applications to calculate the dosage and timing of radioactive tracers.
It assists in nuclear power generation by understanding fuel consumption rates.
Also, it is important in radiation safety to estimate exposure and decay of radioactive waste.
This law helps in determining the age of archaeological and geological samples through radiocarbon dating.
It is used in medical applications to calculate the dosage and timing of radioactive tracers.
It assists in nuclear power generation by understanding fuel consumption rates.
Also, it is important in radiation safety to estimate exposure and decay of radioactive waste.
Explanation:
The law provides a predictable model for how radioactive substances decay over time.
Teachers explain that this predictability allows scientists to use radioactivity as a clock.
It is essential in fields like archaeology, medicine, energy, and environmental science.
Understanding the law helps in practical decision-making regarding radioactive materials.
The law provides a predictable model for how radioactive substances decay over time.
Teachers explain that this predictability allows scientists to use radioactivity as a clock.
It is essential in fields like archaeology, medicine, energy, and environmental science.
Understanding the law helps in practical decision-making regarding radioactive materials.
Question 10
Which of the following is NOT an application of nanotechnology in the automotive industry?
Answer:
c Nanoparticle drug delivery
Explanation:
Nanotechnology applications in the automotive industry include lightweight construction materials, painting with nanofillers to improve durability and appearance, and sensors for vehicle systems to enhance performance and safety.
However, nanoparticle drug delivery is an application related to medicine and not automotive industry.
Therefore, option (c) nanoparticle drug delivery is NOT an application of nanotechnology in the automotive industry.
Nanotechnology applications in the automotive industry include lightweight construction materials, painting with nanofillers to improve durability and appearance, and sensors for vehicle systems to enhance performance and safety.
However, nanoparticle drug delivery is an application related to medicine and not automotive industry.
Therefore, option (c) nanoparticle drug delivery is NOT an application of nanotechnology in the automotive industry.
Question 11
If an electric field of magnitude \(800 \ \mathrm{N\ C^{-1}}\) is applied to a wire, calculate the acceleration of an electron in the wire. (Given \(e = 1.6 \times 10^{-19} \ \mathrm{C}\), \(m = 9.11 \times 10^{-31} \ \mathrm{kg}\))
Answer:
Given electric field \(E = 800 \ \mathrm{N\ C^{-1}}\), charge of electron \(e = 1.6 \times 10^{-19} \ \mathrm{C}\), mass of electron \(m = 9.11 \times 10^{-31} \ \mathrm{kg}\)
Force on electron \(F = eE = 1.6 \times 10^{-19} \times 800 = 1.28 \times 10^{-16} \ \mathrm{N}\)
Acceleration \(a = \frac{F}{m} = \frac{1.28 \times 10^{-16}}{9.11 \times 10^{-31}} = 1.405 \times 10^{14} \ \mathrm{m/s^{2}}\)
Force on electron \(F = eE = 1.6 \times 10^{-19} \times 800 = 1.28 \times 10^{-16} \ \mathrm{N}\)
Acceleration \(a = \frac{F}{m} = \frac{1.28 \times 10^{-16}}{9.11 \times 10^{-31}} = 1.405 \times 10^{14} \ \mathrm{m/s^{2}}\)
Explanation:
The acceleration of an electron in an electric field is found using Newton's second law where the force on the electron is the product of its charge and the electric field. Dividing this force by the electron's mass gives the acceleration.
First, calculate the force using \(F = eE\).
Then, use \(a = \frac{F}{m}\) to find the acceleration.
This approach directly applies the fundamental physics principles of force and acceleration for charged particles in electric fields.
The acceleration of an electron in an electric field is found using Newton's second law where the force on the electron is the product of its charge and the electric field. Dividing this force by the electron's mass gives the acceleration.
First, calculate the force using \(F = eE\).
Then, use \(a = \frac{F}{m}\) to find the acceleration.
This approach directly applies the fundamental physics principles of force and acceleration for charged particles in electric fields.
Question 12
Describe avalanche and Zener breakdown mechanisms.
Answer:
Avalanche breakdown occurs in lightly doped p-n junctions with a wide depletion region. When the reverse bias voltage exceeds a certain value, minority charge carriers gain kinetic energy and collide with atoms, breaking covalent bonds and generating electron-hole pairs. This process multiplies rapidly causing a large reverse current abruptly, called avalanche breakdown.
Zener breakdown occurs in heavily doped p-n junctions with a narrow depletion region. A strong electric field is set up at breakdown voltage which directly breaks covalent bonds, generating electron-hole pairs. This leads to a sharp increase in reverse current called Zener breakdown.
Zener breakdown occurs in heavily doped p-n junctions with a narrow depletion region. A strong electric field is set up at breakdown voltage which directly breaks covalent bonds, generating electron-hole pairs. This leads to a sharp increase in reverse current called Zener breakdown.
Explanation:
Avalanche breakdown is explained by the acceleration of minority carriers in a wide depletion region, causing collisions and multiplication of charge carriers.
Zener breakdown happens due to a strong electric field in a narrow depletion region breaking bonds directly.
Understanding the doping level and depletion width helps distinguish these mechanisms.
This explanation helps students visualize physical processes behind breakdowns in diodes.
Avalanche breakdown is explained by the acceleration of minority carriers in a wide depletion region, causing collisions and multiplication of charge carriers.
Zener breakdown happens due to a strong electric field in a narrow depletion region breaking bonds directly.
Understanding the doping level and depletion width helps distinguish these mechanisms.
This explanation helps students visualize physical processes behind breakdowns in diodes.
Question 13
Which statement best describes gamma emission from an excited nucleus?
Answer:
Gamma emission is the process by which an excited nucleus releases excess energy by emitting a gamma photon (\(\gamma\)) without changing its atomic or mass number.
Explanation:
Gamma emission involves the release of electromagnetic radiation (gamma rays) from an excited nucleus.
Unlike alpha or beta decay, gamma emission does not change the number of protons or neutrons, so the element remains the same.
It is a way for the nucleus to lose excess energy and reach a more stable state.
Gamma emission involves the release of electromagnetic radiation (gamma rays) from an excited nucleus.
Unlike alpha or beta decay, gamma emission does not change the number of protons or neutrons, so the element remains the same.
It is a way for the nucleus to lose excess energy and reach a more stable state.
Question 14
Calculate the momentum and de Broglie wavelength of a bullet of mass 50 g moving at 200 m/s and explain why its wave nature is not observable.
Answer:
Step 1 Convert mass to kg: 50 g = 0.05 kg
Step 2 Calculate momentum \( p = mv = 0.05 \times 200 = 10 \ \text{kg m/s} \)
Step 3 Calculate de Broglie wavelength \( \lambda = \frac{h}{p} = \frac{6.626 \times 10^{-34}}{10} = 6.626 \times 10^{-35} \ \text{m} \)
Step 4 Explanation: The wavelength is extremely small (on the order of \( 10^{-35} \ \text{m} \)), which is far smaller than atomic dimensions and cannot be detected by any experimental apparatus. Hence, the wave nature of macroscopic objects like bullets is not observable.
Step 2 Calculate momentum \( p = mv = 0.05 \times 200 = 10 \ \text{kg m/s} \)
Step 3 Calculate de Broglie wavelength \( \lambda = \frac{h}{p} = \frac{6.626 \times 10^{-34}}{10} = 6.626 \times 10^{-35} \ \text{m} \)
Step 4 Explanation: The wavelength is extremely small (on the order of \( 10^{-35} \ \text{m} \)), which is far smaller than atomic dimensions and cannot be detected by any experimental apparatus. Hence, the wave nature of macroscopic objects like bullets is not observable.
Explanation:
First, calculate the momentum of the bullet using classical mechanics \( p = mv \). Then use the de Broglie formula to find the wavelength. The resulting wavelength is extraordinarily small compared to atomic scales, making any wave-like behavior undetectable. This illustrates why wave nature is significant only at microscopic scales, such as electrons, and not for everyday macroscopic objects.
First, calculate the momentum of the bullet using classical mechanics \( p = mv \). Then use the de Broglie formula to find the wavelength. The resulting wavelength is extraordinarily small compared to atomic scales, making any wave-like behavior undetectable. This illustrates why wave nature is significant only at microscopic scales, such as electrons, and not for everyday macroscopic objects.
Question 15
Calculate the speed of an electromagnetic wave in a medium where the electric field amplitude is \(1.2 \times 10^4 \, \text{N C}^{-1}\) and the magnetic field amplitude is \(4 \times 10^{-5} \, \text{T}\).
Answer:
Given:
Electric field amplitude, \(E_0 = 1.2 \times 10^4\) N C\(^{-1}\)
Magnetic field amplitude, \(B_0 = 4 \times 10^{-5}\) T
The speed of the electromagnetic wave in the medium is
\[ v = \frac{E_0}{B_0} = \frac{1.2 \times 10^4}{4 \times 10^{-5}} = 3 \times 10^8 \text{ m/s} \]
Therefore, the speed of the electromagnetic wave in the medium is \(3 \times 10^8\) m/s.
Electric field amplitude, \(E_0 = 1.2 \times 10^4\) N C\(^{-1}\)
Magnetic field amplitude, \(B_0 = 4 \times 10^{-5}\) T
The speed of the electromagnetic wave in the medium is
\[ v = \frac{E_0}{B_0} = \frac{1.2 \times 10^4}{4 \times 10^{-5}} = 3 \times 10^8 \text{ m/s} \]
Therefore, the speed of the electromagnetic wave in the medium is \(3 \times 10^8\) m/s.
Explanation:
The speed of an electromagnetic wave in a medium is the ratio of the electric field amplitude to the magnetic field amplitude.
By dividing the given electric field amplitude by the magnetic field amplitude, we find the wave speed.
This formula is derived from Maxwell's equations and is fundamental in understanding wave propagation in different media.
The speed of an electromagnetic wave in a medium is the ratio of the electric field amplitude to the magnetic field amplitude.
By dividing the given electric field amplitude by the magnetic field amplitude, we find the wave speed.
This formula is derived from Maxwell's equations and is fundamental in understanding wave propagation in different media.
Question 16
When a metallic surface is illuminated with radiation of wavelength \( \lambda \), the stopping potential is \( V \). If the same surface is illuminated with radiation of wavelength \( 2\lambda \), the stopping potential is \( \frac{V}{4} \). The threshold wavelength for the metallic surface is
Answer:
Explanation:
Using the photoelectric equation relating stopping potential and wavelength, we set up two equations for the two wavelengths. By algebraic manipulation, we find the work function and then the threshold wavelength. The threshold wavelength is found to be three times the original wavelength \( \lambda \). This shows how stopping potential changes with wavelength and helps find the threshold wavelength of the metal.
Using the photoelectric equation relating stopping potential and wavelength, we set up two equations for the two wavelengths. By algebraic manipulation, we find the work function and then the threshold wavelength. The threshold wavelength is found to be three times the original wavelength \( \lambda \). This shows how stopping potential changes with wavelength and helps find the threshold wavelength of the metal.
Question 17
Two capacitors of capacitances \( C_1 \) and \( C_2 \) are connected in series. The equivalent capacitance \( C_s \) is given by
Answer:
Explanation:
When capacitors are connected in series, the charge on each capacitor is the same but the voltage divides among them. The total voltage across the series combination is the sum of voltages across each capacitor. Using the relation \( Q = CV \), and the fact that charge \( Q \) is same on each capacitor, the equivalent capacitance is found using the reciprocal formula. This is a fundamental formula for capacitors in series and is often memorized by students.
When capacitors are connected in series, the charge on each capacitor is the same but the voltage divides among them. The total voltage across the series combination is the sum of voltages across each capacitor. Using the relation \( Q = CV \), and the fact that charge \( Q \) is same on each capacitor, the equivalent capacitance is found using the reciprocal formula. This is a fundamental formula for capacitors in series and is often memorized by students.
Question 18
Calculate the energy stored in an inductor of inductance 0.5 H carrying a current of 4 A.
Answer:
Given inductance L = 0.5 H and current i = 4 A.
Energy stored in the inductor is given by the formula
U = (1/2) L i^2
Substitute values:
U = (1/2) × 0.5 × (4)^2 = 0.25 × 16 = 4 J
Therefore, the energy stored in the inductor is 4 joules.
Energy stored in the inductor is given by the formula
U = (1/2) L i^2
Substitute values:
U = (1/2) × 0.5 × (4)^2 = 0.25 × 16 = 4 J
Therefore, the energy stored in the inductor is 4 joules.
Explanation:
The energy stored in an inductor is the magnetic potential energy due to the current flowing through it.
The formula U = (1/2) L i^2 directly relates inductance and current to the stored energy.
By substituting the given values, the calculation is straightforward.
This formula is derived from the work done to establish the current against the self-induced emf.
The energy stored in an inductor is the magnetic potential energy due to the current flowing through it.
The formula U = (1/2) L i^2 directly relates inductance and current to the stored energy.
By substituting the given values, the calculation is straightforward.
This formula is derived from the work done to establish the current against the self-induced emf.
Question 19
Derive the expression for the horizontal component of Earth's magnetic field \(B_H\) using the tangent law and the magnetic field produced by the coil in a tangent galvanometer.
Answer:
Let \(B\) be the magnetic field due to the coil and \(B_H\) be the horizontal component of Earth's magnetic field.
According to tangent law:
\[ B = B_H \tan \theta \quad (1) \]
The magnetic field at the centre of the coil is:
\[ B = \frac{\mu_0 N I}{2 R} \quad (2) \]
Equate (1) and (2):
\[ \frac{\mu_0 N I}{2 R} = B_H \tan \theta \]
Rearranging for \(B_H\):
\[ B_H = \frac{\mu_0 N I}{2 R \tan \theta} \]
According to tangent law:
\[ B = B_H \tan \theta \quad (1) \]
The magnetic field at the centre of the coil is:
\[ B = \frac{\mu_0 N I}{2 R} \quad (2) \]
Equate (1) and (2):
\[ \frac{\mu_0 N I}{2 R} = B_H \tan \theta \]
Rearranging for \(B_H\):
\[ B_H = \frac{\mu_0 N I}{2 R \tan \theta} \]
Explanation:
The derivation starts from the tangent law which relates the coil's magnetic field to Earth's horizontal magnetic field and the deflection angle. Using the known formula for the coil's magnetic field, equate the two expressions and solve for \(B_H\). This derivation helps students connect the physical setup with the mathematical expression and understand how \(B_H\) can be experimentally determined.
The derivation starts from the tangent law which relates the coil's magnetic field to Earth's horizontal magnetic field and the deflection angle. Using the known formula for the coil's magnetic field, equate the two expressions and solve for \(B_H\). This derivation helps students connect the physical setup with the mathematical expression and understand how \(B_H\) can be experimentally determined.
Question 20
If two conducting spheres of radii r1 and r2 are connected by a wire and carry charges q1 and q2 respectively, which of the following relations holds true at electrostatic equilibrium?
Answer:
Explanation:
At electrostatic equilibrium, the potential on both spheres is equal.
Potential on a sphere is given by \(V = \frac{1}{4 \pi \epsilon_0} \frac{q}{r}\).
Equating potentials, \(\frac{q_1}{r_1} = \frac{q_2}{r_2}\) holds true.
This relation is fundamental in understanding charge distribution on connected conductors.
At electrostatic equilibrium, the potential on both spheres is equal.
Potential on a sphere is given by \(V = \frac{1}{4 \pi \epsilon_0} \frac{q}{r}\).
Equating potentials, \(\frac{q_1}{r_1} = \frac{q_2}{r_2}\) holds true.
This relation is fundamental in understanding charge distribution on connected conductors.
